bạn cần giúp bài nào ạ?

a) \(5\sqrt{12}+4\sqrt{27}-6\sqrt{48}=10\sqrt{3}+12\sqrt{3}-24\sqrt{3}=-2\sqrt{3}\)

b) \(\left(\sqrt{300}-2\sqrt{675}+5\sqrt{75}\right):\sqrt{3}=\left(10\sqrt{3}-30\sqrt{3}+25\sqrt{3}\right):\sqrt{3}\)

\(=10-30+25=5\)

d) \(\dfrac{2}{4-2\sqrt{3}}+\dfrac{2}{4+2\sqrt{3}}=\dfrac{1}{2-\sqrt{3}}+\dfrac{1}{2+\sqrt{3}}\)

\(=\dfrac{2+\sqrt{3}+2-\sqrt{3}}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}=4\)

Bài 4: ĐK: x>0

a)  \(B=\dfrac{x^2+\sqrt{x}}{x-\sqrt{x}+1}+1-\dfrac{2x+\sqrt{x}}{\sqrt{x}}\)

\(\Leftrightarrow B=\dfrac{\sqrt{x}\left[\left(\sqrt{x}\right)^3+1\right]}{x-\sqrt{x}+1}+1-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}\)

\(\Leftrightarrow B=\dfrac{\sqrt{x}.\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}+1-2\sqrt{x}-1\)

\(\Leftrightarrow B=\sqrt{x}.\left(\sqrt{x}+1\right)-2\sqrt{x}=x+\sqrt{x}-2\sqrt{x}\)

\(\Leftrightarrow B=x-\sqrt{x}\)

Vậy với x>0 thì \(B=x-\sqrt{x}\)

b) Ta có: \(B=2\)

\(\Leftrightarrow x-\sqrt{x}=2\)

\(\Leftrightarrow x-\sqrt{x}-2=0\)

 \(\Leftrightarrow x-2\sqrt{x}+\sqrt{x}-2=0\)

\(\Leftrightarrow\sqrt{x}.\left(\sqrt{x}-2\right)+\left(\sqrt{x}-2\right)=0\) 

\(\Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)=0\)

Do \(\sqrt{x}+1>0\) nên, ta suy ra:

\(\sqrt{x}-2=0\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\) \(\left(TMĐK\right)\)

Vậy \(x=4\) thì \(B=2\)

C38. 1087110 hay 1078110 thế ạ??

Bài 8: 

\(P=5\sqrt{5}+2\sqrt{5}-6\sqrt{5}=\sqrt{5}\)

a) \(A=15\sqrt{3}-5\sqrt{3}-4\sqrt{3}=6\sqrt{3}\)

b) \(B=\dfrac{\sqrt{3}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}-\dfrac{\sqrt{3}\left(\sqrt{5}+1\right)}{\sqrt{5}+1}=\sqrt{3}-\sqrt{3}=0\)

a) \(A=10\sqrt{2}-12\sqrt{2}-9\sqrt{2}+35\sqrt{2}=24\sqrt{2}\)

b) \(B=\sqrt{\left(\sqrt{7}+1\right)^2}+\sqrt{\left(\sqrt{7}-2\right)^2}\)

\(=\sqrt{7}+1+\sqrt{7}-2=2\sqrt{7}-1\)

c) \(C=2\sqrt{5}+\dfrac{8\left(3-\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}-\dfrac{3\left(\sqrt{2}-\sqrt{5}\right)}{\left(\sqrt{2}-\sqrt{5}\right)}\)

\(=2\sqrt{5}+\dfrac{24-8\sqrt{5}}{4}-3=2\sqrt{5}+6-2\sqrt{5}-3=3\)

refused - would have